-16t^2=128t

Solution for -16t^2=128t equation:

-16t^2=128t

We move all terms to the left:

-16t^2-(128t)=0

a = -16; b = -128; c = 0;
Δ = b2-4ac
Δ = -1282-4·(-16)·0
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-128}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+128}{2*-16}=\frac{256}{-32}
=-8
$